Efficiency Of Solar Cells

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Yes, solar panels can be installed on hospitals or healthcare facilities. In fact, many healthcare facilities are increasingly adopting solar energy as a sustainable and cost-effective solution. Solar panels can help hospitals reduce their dependence on traditional energy sources, lower operating costs, and contribute to a greener environment.
Yes, solar panels can be used in remote areas with no access to the grid. Solar panels generate electricity from sunlight, making them an ideal solution for off-grid locations where traditional power infrastructure is unavailable. They can be installed in remote areas to provide a reliable and sustainable source of energy, allowing for essential power needs such as lighting, charging devices, and running small appliances even in areas far from the grid.
i don't have a solar panel yet so can you tell me what votage panel should i get. do i need an inverter or something.
Solar panels can greatly enhance a property's energy independence by generating electricity from sunlight, thus reducing reliance on traditional energy sources. This allows homeowners or businesses to generate their own clean and renewable energy, potentially offsetting or even eliminating their need for grid electricity.
i dont wanna buy a solar panel becuase its too exspensive, i dont want use technical stuff i want to use every day materiels, but i need to know asap plzzzz.
Solar panel is a sheet of highly expensive, purest Silicon manufactured in a factory of exceptional killingly stringent specifications. You can't do that as a cottage industry, at least the material.
I am doing research for a project.Please reply soon! :)
Photolvoltaic arry is the more technical name and the more accurate name for solar panels. Photo indicates light and voltaic indicates a voltage is produced.. Array is just a layout of these little light sensitive electricity generating panels. Actually Solar only idicates the SUN and not light in itself..
I am planning to run a couple of dc motors on solar current. I have a couple of 450mA 4V, 200mA .5V, 00mA 9V solar panels with me. I need to run a pair of .5A 9V motors. Suggest me a circuit along with other equipments I might require. For your info, I am trying to build a solar powered drone / UAV
I don't think it is enough solar power. Your motor requires .5A x 9V = 3.5 watts. The solar cells are .450A x 4V x 2ea. = 3.6 watts, .2A x .5V = .3 watts, .A x 9V = .9 watts total solar = 4.8 watts. The problem with connecting the cells in series to get the right voltage, like the two 4V cells in series with the .5 volt cell (total 9.5 volts) is that the .5 volt cell limits the maximum current to only 0.2A. If you paralleled that with the 9V cell, you would get 9 volts at 0.2 + 0. = 0.3A, too low to operate the motor at its design current and voltage. The best you could do is put the two 4V cells in series, and in parallel with the 9V cell (a blocking diode might be a good idea, but probably not required for such a small array). That way you get 8 volts at 0.55 amps.
If my school uses 88240kWh of electricity per month and I have 000 50W solar panels running for 6 hours a day, does it mean it will take 88240kWh / {{[(50W x 3600s)*000]/000}kWh x 6} number of hours to generate that much electricity (88240kWh)?
running six hours a day doesn't mean much. You need to look at the solar insulation charts for your school's geographical location to come up with a better factor. The easiest number for you to use is sun hours. For example, Washington DC averages 4.23 hours. Do a Yahoo search for sun hours and you should find lots of charts. Solar panels rated at 50W give this output at full sun near noon at full brightness (no clouds). The sun hour factor makes it easy to find the equivalent number of full brightness hours. So, using Washington DC as an example you have: 000 panels * 50W * 4.23 sunhours/day = 634kWh a day on average. You state your school uses 88240kWh/month which is 6274kWh a day. This would mean you need ten times more solar panels since there is no way to get more daylight. Be careful to put in all the units in your formula and cancel them out to make sure you don't end up with a nonsense result. The title of the question would be answered as followed: 50W/000 * 4.23 sunhours/day = 0.63kWh/day or 9kWh per month or 228kWh a year. These are annual averages. If you wanted a specific month, you would need the sun hours for that month. Hope this helps.